A Zener diode in the circuit shown has a knee | vedclass

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(C) The total current $I$ supplied by the source is given by Ohm's law applied to the series resistor:$I = \frac{V_{in} - V_z}{R} = \frac{9\,V - 6\,V}

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$10\,mA, 55\,mA$

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(C) The total current $I$ supplied by the source is given by Ohm's law applied to the series resistor:$I = \frac{V_{in} - V_z}{R} = \frac{9\,V - 6\,V}{50\,\Omega} = \frac{3\,V}{50\,\Omega} = 0.06\,A = 60\,mA$.The Zener diode must operate between its knee current $I_{z,min} = 5\,mA$ and its maximum current $I_{z,max}$ determined by the power dissipation limit $P_{max} = 300\,mW$.$I_{z,max} = \frac{P_{max}}{V_z} = \frac{300\,mW}{6\,V} = 50\,mA$.The load current $I_L$ is given by $I_L = I - I_z$.To find the maximum load current $I_{L,max}$, we use the minimum Zener current:$I_{L,max} = I - I_{z,min} = 60\,mA - 5\,mA = 55\,mA$.To find the minimum load current $I_{L,min}$, we use the maximum Zener current:$I_{L,min} = I - I_{z,max} = 60\,mA - 50\,mA = 10\,mA$.Thus, the minimum and maximum load currents are $10\,mA$ and $55\,mA$ respectively.

$A$ Zener diode in the circuit shown has a knee current of $5\,mA$ and a maximum allowed power dissipation of $300\,mW$. What are the minimum and maximum load currents that can be drawn safely from the circuit, keeping the output voltage $V_0$ constant at $6\,V$?

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